The discussion revolves around solving a partial differential equation, specifically the Laplace equation in two dimensions, expressed as \(\frac{{\partial}^{2}f}{{\partial x}^{2}}+\frac{{\partial}^{2}f}{{\partial y}^{2}}=0\). Participants explore various aspects of the general solution, including the role of boundary conditions and the nature of the solutions.
Participants express differing views on the necessity of boundary conditions and the implications of the general solution. While there is a shared understanding of the form of the solution, the discussion does not reach a consensus on the implications of the boundary conditions or the nature of the solutions without them.
The discussion highlights the dependence of solutions on boundary conditions and the implications of analyticity versus differentiability in the context of the Laplace equation.
jackmell said:The general solution is:
[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]
where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:
[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]
where:
[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]
So first solve:
[tex]u_t-iu_x=0[/tex]
which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:
[tex]f_x+if_y=g(x+it)[/tex]
another easy one to obtain so that the general solution is:
[tex]f(x,y)=g(x+it)+h(x-it)[/tex]
Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.